In a race among 10 contestants, how many orders of finish are there, counting possibilities of ties?

For example, in a 4-person race, A and B finishing at the same time, before C and D finish simultaneously, is a different order from C and D finishing together before A and B finishing together, and of course each of the 4! ways of all finishing separately are different orders.

(In reply to simple computation algorithm by Daniel)

Daniel:

Nice work.  However, I think what you meant to say is

... then for every kth element with k<N we get its value by adding the k and k-1 value from the N-1 list AND MULTIPLYING THE SUM BY k

At least, that what I think your sample indicates

Posted by Steve Herman on 2008-01-13 14:29:36